How Come In This Code To Find Prime Numbers, Is_prime(9) Returns True?

def is_prime(x): if x < 2: return False else: for n in range(2, x): if x % n == 0: return False else: return True print is_prime(9) ret

Solution 1:

This is because you don't actually loop, as you return True during the first cycle (9 % 2 == 0 is False).

Something like this should solve the problem:

defis_prime(x):
  if x < 2:
    returnFalsefor n inrange(2, x):
    if x % n == 0:
      returnFalsereturnTrue

Solution 2:

You can simplify the logic a good amount by keeping your original loop and not exiting early. You can add your first conditional to your final return:

defis_prime(x):
  for n inrange(2, x):
    if x % n == 0:
      returnFalsereturn x > 2

BTW, the Sieve of Erastothenes is a pretty cool method of solving this problem in a much better run time complexity. Here's a link to a brief explanation:

https://math.stackexchange.com/questions/58799/why-in-sieve-of-erastothenes-of-n-number-you-need-to-check-and-cross-out-numbe